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</p>
<p>Wow! Do you happen to learn vectors before number systems; and integration before differentiation? I actually like the education system you’re describing here. I find differential equations and integrals more intuitive than trigonometry, too. But I think it’s hard to get a complete treatment of the former two without knowing the latter.</p>
<p>Sorry to hear that you’re disadvantaged on the SAT II Math.</p>
<p>We learn the first two mentioned at physics. What do you mean by integration BEFORE differentiation? </p>
<p>@MrPropapanda Wow are you really 15. Your profile says so.</p>
<p>Oh, then SAT II Physics shouldn’t be a huge problem for you. Mmm, I meant as in, were you taught a chapter on integration before you were taught a chapter on differentiation? Most people learn differentiation first.</p>
<p>
:)</p>
<p>@blackhair1
How do you guys solve the equations like y"=-y or some thing like y’=1+y^2</p>
<p>I was wondering that too. So much of trig shows up in calculus</p>
<p>I am not sure what you want me to solve.
The first one. I am not sure what you want. </p>
<p>y’‘=-y
y’‘’=-1
y’= -(1/2)y^2 + c (any number without an letter)
y = - (1/6)y^3 + c (any number without an letter)</p>
<p>the second one
I am guessing you want the integral of that function
y’=1+y^2
y = y + (1/3)y^3 + c (any number without an letter)
or you want the diffentation?
y’'=2y</p>
<p>I do not feel like bragging, but you may also ask me to differentiate something with e or fractions. If SAT II had these kind of questions, I would be very happy.</p>
<p>Maybe they use exponentials with complex powers (trig functions without the ‘sin’, ‘cosine’ etc.)?</p>
<p>No, he was asking to solve for ‘y’, as a function (find the function). What you’re doing is differentiating/integrating the y derivatives… taking the ‘y’ on the RHS to mean the variable (y(y) O_o)?</p>
<p>Have you done ‘differential equations’, or just ‘differentiation’?</p>
<p>Uh… I am getting lost in translation.</p>
<p>Lets say you have x’’ + b/m<em>x’ + k/m</em>x = 0 (the usual oscillation + damping thing). Here, you have to find x as a function of time (find the expression for x(t)), not just simply ‘differentiate’ or ‘integrate’.</p>
<p>What’s usually done is take x = Ae^(a*t) or such, and mess around (not sure about the specifics, we haven’t done this in school yet (grade 11 
)).</p>
<p>y"=-y, y’=1+y^2 </p>
<p>Do you mean this?
y"=-y
y=-y"</p>
<p>y’=1+y^2
1+y^2 =y’
y^2 =y’ - 1
y = the root of (y’ - 1)</p>
<p>No, you’re just manipulating the expression. Like, to solve ‘x = 2x - 1’, you’re doing,</p>
<p>2x = 3x - 1
0 = 2x - x - 1</p>
<p>And such.</p>
<p>You have to find y, not in terms of itself or its derivatives (if I understood the asker correctly).</p>
<p>Not exactly… differential equations are when your variables are derivatives. So:</p>
<p>y’=1+y^2 … y’=dy/dx</p>
<p>dy/dx=1+y^2 … Seperate variables</p>
<p>dy/(1+y^2)=dx … Integrate both sides</p>
<p>so dx – > x +c</p>
<p>but to integrate dy/(1+y^2) you need arctan (as far as i know)</p>
<p>dy/(1+y^2)–> arctan(y) +c</p>
<p>So:
arctan(y)=x+c</p>
<p>Someone check my math, it’s pretty late here</p>
<p>Well, thats what I meant. :P</p>
<p>For y’’ = -y, y = A*sin(x) would do right?</p>
<p>Isn’t there also a Bcos(x) somewhere there as well? Plus or minus, not sure</p>
<p>lol, first how do you learn differential equations without trig, how would you solve 2y’'+3y=cos(x). and there’s numerous ways of solving differential equations, the easiest being separating (getting the dy’s on one side, and dx’s on the other) then if the d.e is nice you can use the characteristic equation, and so on…this gets more complicated with Frobenius method and Laplace transforms and so on…</p>
<p>A good method is… Guessing. :P</p>
<p>I think he’s just done ‘differentiation’, not ‘differential equations’.</p>
<p>Anyway, you can solve ‘trig-ish’ equations with complex exponentials, and no trig functions.</p>
<p>@blackhair1 Sorry for not making my question clear.</p>
<p>Consider the equation y=y’ a simple solution for this will be exponential function y=e^x since derivative of e^x is it self. (e^x)=(e^x)'</p>
<p>A particular solution for y"=-y is y=sin(x). The trig function sin(x) has such property, (sin(x))‘=(cos(x)) and (cos(x))’=-sin(x) therefore (sin(x))“=-sin(x) or y”=-y. The other equation was already solved by Confused92. </p>
<p>
May be,
@blackhair1, if you are familiar with the complex powers then sin(x)=(e^(ix)-e^(-ix))/(2i) and cos(x)=(e^(ix)+e^(-ix))/(2i).</p>
<p>I understand you because I did not get good education too. So don’t worry, hopefully MIT understands your case. You don’t need differential equations to get into MIT. But knowing little trig may help, so PM me and I can send you some useful links where you can self study trig functions.</p>
<p>Yea, then I probably have done differentiation. I thought it meant the same
I don’t believe i don’t got good education here, it is just different from what you learn. Now i understand you. My school has never focused on differentials with trigonometry.</p>
<p>Don’t worry too much btw. I hoped that the admission rate for international applicants is lower due to lower SAT scores. If you have some useful links for me, I will greatly appreciate that. Please pm me.</p>
<p>Inconclusive, may I ask you from which country you are?</p>
<p>I am Somali pirate! 
I am US permanent resident though, not international applicant.
I will.</p>